CDF approximation - L'Hopital's rule

We have that

$$left( frac{ (1-phi) x}{phi (k-1)}+1right)^ {1-k}=e^{(1-k)logleft( frac{ (1-phi) x}{phi (k-1)}+1right)}$$

and

$$(1-k)logleft( frac{ (1-phi) x}{phi (k-1)}+1right)=-frac{logleft( frac{ (1-phi) x}{phi (k-1)}+1right)}{frac 1{k-1}}=\=-frac{ (1-phi) x}{phi }frac{logleft( frac{ (1-phi) x}{phi (k-1)}+1right)}{frac{ (1-phi) x}{phi (k-1)}}to frac{ (1-phi) x}{phi }$$

since, by standard limits, as $t to 0$ we have $frac{log (1+t)}{t} to 1$.

Note that it suffices to approximate

$$left(1+frac{(phi-1)x/phi}{1-k}right)^{1-k}$$

by $e^{frac{(phi-1)x}{phi}}$.

With the substitution $n=1-k$ and $r=(phi-1)x/phi$, this follows immediately from the fact that

$$lim_{nto-infty}left(1+frac{r}{n}right)^n=e^r$$

You can prove this by L'Hospital applied to $frac{ln(1+rz^{-1})}{z^{-1}}$. Note that differentiation on top and bottom gives

$$frac{r}{1+rz^{-1}}$$

which tends to $r$ as $zto-infty$.